To solve the problem and find the maximum number of ride tickets [tex]\( r \)[/tex] Alana can buy, let's follow a structured approach:
1. Define the variables:
- Let [tex]\( r \)[/tex] represent the number of ride tickets.
- Let [tex]\( f \)[/tex] represent the number of food tickets.
2. Set up the inequalities based on the problem description:
- Alana needs to buy at least a total of 16 tickets:
[tex]\[
r + f \geq 16
\][/tex]
- Alana has at most \[tex]$40 to spend where ride tickets cost \$[/tex]4 each and food tickets cost \$2 each:
[tex]\[
4r + 2f \leq 40
\][/tex]
3. Simplify the inequalities:
- Simplify the second inequality by dividing through by 2:
[tex]\[
2r + f \leq 20
\][/tex]
4. Consider the boundary conditions:
- To find the feasible region, let's convert the inequalities to equations:
[tex]\[
r + f = 16
\][/tex]
[tex]\[
2r + f = 20
\][/tex]
5. Solve the system of equations:
- Set the equations equal to each other by substituting [tex]\( f \)[/tex] from the second equation into the first:
[tex]\[
f = 20 - 2r
\][/tex]
Substitute [tex]\( f \)[/tex] into the first equation:
[tex]\[
r + (20 - 2r) = 16
\][/tex]
Simplify and solve for [tex]\( r \)[/tex]:
[tex]\[
r + 20 - 2r = 16
\][/tex]
[tex]\[
-r + 20 = 16
\][/tex]
[tex]\[
r = 4
\][/tex]
6. Verify the solution:
- Substitute [tex]\( r = 4 \)[/tex] back into one of the original inequalities to ensure it satisfies both conditions:
[tex]\[
r + f = 4 + f \geq 16 \quad \text{implies} \quad 4 + f = 16 \quad \text{so}\quad f = 12
\][/tex]
[tex]\[
2r + f = 2 \times 4 + 12 = 8 + 12 = 20 \quad \text{(which is also less than 20)}
\][/tex]
Thus, the maximum number of ride tickets Alana can buy is indeed [tex]\( 4 \)[/tex].
So, the answer is [tex]\(\boxed{4}\)[/tex].
